# how come square root of 2 times itself equals 2

since square root of $$2$$ is a irrational number, which we know or assume its decimal part is not a finite number, or doesn't terminate, how come we say that this infinite number (not in terms of being a large number) times itself is equals $$2$$. It is like saying $$0.928041\dots$$ times itself is some finite number (like $$1.64$$ not going $$3$$ dots in the end).

• If it helps, you can represent $2$ by $2.000\ldots$ They're two ways of saying the same thing. – Theo Bendit Feb 20 at 10:09
• Isn't that the definition of the square root? "A number which produces a specified quantity when multiplied by itself." – Vepir Feb 20 at 10:10
• Note that $0.9999\ldots\cdot0.9999\ldots=1\cdot1=1$. – Brian M. Scott Feb 20 at 10:15

## 3 Answers

There is no such thing as an infinite number. If you mean that the decimal expansion of $$\sqrt2$$ never ends, then you are right. But that also applies to the decimal expansion of $$\frac13$$; however, I suspect that you see no problem with the assertion that $$3\times\frac13=1$$.

The fact that the decimal expansion of $$\sqrt2$$ never ends is something that concerns the specific way that we use to represent numbers, which was developed centuries after square roots started to be studied. It is not something about the number $$\sqrt2$$ itself, which is defined as the only real number greater than or equal to $$0$$ whose square is $$2$$.

After having defined it this way, we can try to compute its decimal expansion. For instance, $$1^2<2<2^2$$, and therefore $$1<\sqrt2<2$$. Furthermore, $$\left(\frac75\right)^2<2<\left(\frac85\right)^2$$ and therefore $$\frac75<\sqrt2<\frac85(1.4<\sqrt2<1.6)$$.

• derivation of this "a=0.3333... then 10a=3.333.... then a=1/3 " is weird too. how come you multiply something that never ends and then subtract it with another thing that also never ends. So I don't really agree on 0.3333... = 1/3. – ozgun can Feb 20 at 10:28
• That is just not something you can disagree with. One can use calculus to prove $1/3=0.333...$. That is not a matter of opinion. – Lex_i Feb 20 at 10:32
• You are mixing two things that you should not mix: the number $\frac13$ and its decimal representation $0.3333333\ldots$. We define $\frac13$ in such a way that $3\times\frac13=1$. And when we compute its decimal representation we get $0.3333333\ldots$ – José Carlos Santos Feb 20 at 10:32
• @ozguncan $0.333...$ can be expressed as the infinite sum $\sum_{n=1}^{\infty}3(1/10)^n$. This is a geometric sum that equals exactly $1/3$ – Lex_i Feb 20 at 10:41

It's by definition. $$\sqrt 2$$ is the unique positive number that, when squared, results in $$2$$. Are you also surprised that $$\frac13+\frac13+\frac13=1$$ even though you are adding a number repeatedly to itself that has a decimal representation that is non-terminating? (Similar to above, $$\frac13$$ is defined as the unique number that, when multiplied by $$3$$, results in $$1$$)

Do not confuse a number with the representation of a number by a decimal expansion.

• The last sentence is probably good advice. But. It is possible (and maybe even traditional) to construct the real numbers as decimals. I remember that when I first taught analysis (c. 1965) there had been set the previous year as a finals examination question "Explain briefly how to construct the real numbers by Dedekind Cuts, Cauchy Sequences, Decimals. Prove that these yield isomorphic ordered fields." – ancientmathematician Feb 20 at 10:20

In the real world, you only want to be accurate up to a point. Within an inch, or a millimetre, or much less if you are building a computer chip.
Two people might agree that two numbers are close enough to being equal, but a third might disagree.
If two numbers are so close that everyone agrees they are close enough - so they agree to every decimal point - then we say they are equal. So $$1/3=0.333333333...$$